Show $\displaystyle{\sum_{k=1}^\infty (-1)^{k} x^{\sqrt{2}+k } }$ converges uniformly on $[0,b]$ for $b$ less than $1$.
The series converges by alternating series test, since the term $x^{\sqrt{2}+k}$ goes to zero in the interval,along with being monotonically decreasing (except at zero).
But l don't think it shows uniform convergence. Which test should l use to prove uniform convergence ? The Weierstrass M test and Abel's test don't seem to work.
I am not a hundred percent sure, but the fastest way is to use total convergence.
In fact total convergence $\implies$ uniform convergence. \begin{equation} \sum_{k=1}^\infty \left|(-1)^{k} x^{\sqrt{2}+k}\right|=\sum_{k=1}^\infty \left| x^{\sqrt{2}}x^k\right|=x^{\sqrt{2}}\sum_{k=1}^\infty \left| x^k\right| \end{equation} Now, we now by our hypothesis that $x \in [0,b]$ so we can say $x^k \leq b^k \ \forall x \in [0,b]$ and $\forall k \in \mathbb{N}$. Therefore \begin{equation} \sum_{k=1}^\infty \left| x^k\right| \leq \sum_{k=1}^\infty \left| b^k\right|=\sum_{k=1}^\infty b^k \end{equation} Since $\displaystyle{|b| <1, \ \sum_{k=1}^\infty b^k}$ converges because it is a geometric series. It follow that the original series converges uniformly in the given interval